Chem Labs uses raw materials I and II to produce two domestic cleaning solutions, A and B. The daily availabilities of raw materials I and II are 150 and 145 units, respectively. One unit of solution A consumes 0.5 unit of raw material I and 0.6 unit of raw material II, and one unit of solution B uses 0.5 unit of raw material I and 0.4 unit of raw material II. The profits per unit of solutions A and B are 8 and $10, respectively. The daily demand for solution A lies between 30 and 150 units, and that for solution B between 40 and 200 units. Find the optimal production amounts of A and B.
My attempt
Let A and B be the no. of units of A and B produced and X and Y be no. of raw materials I and II to be processed respectively.
The objective function is to maximize the profit, Z.
Z=8A+10B
The objective function is subject to the following constraints
30<=0.5X+0.6Y<=150
40<=0.5X+0.4Y<=200
X<=150
Y<=145
Is this formulation correct? If it is, how can one proceed from this point to find the maximum profit?
The use of raw materials $I$ and $II$ depends on the production of $A$ and $B$, so you don't need the variables $X$ and $Y$.
You can create a table of given data:
$$\begin{array}{c|c|c|c} Products&I&II&Profit&Demand\\ \hline A&0.5&0.6&8&30\le A\le 150\\ B&0.5&0.4&10&40\le B\le 200\\ \hline Available&\le 150&\le 145&maximize&\end{array}$$
Now we can formulate the LPP: let $A$ and $B$ be the numbers of units of $A$ and $B$, respectively. Then: $$\pi(A,B)=8A+10B\to \text{max} \ \ \text{subject to}\\ 0.5A+0.5B\le 150 \ \ \text{(material I constraint)}\\ 0.6A+0.4B\le 145 \ \ \text{(material II constraint)}\\ 30\le A\le 150 \ \ \text{(demand for A)}\\ 40\le B\le 200 \ \ \text{(demand for B)}\\ $$ You can use graphical or Simplex methods to solve LPP.
Graphical method.
1) Draw the feasible (green) region from the constraint inequalities:
2) Find the corner points: $A,B,C,D,E,F$.
3) Evaluate the objective (profit) function at the corner points and choose the maximum.
Can you do it?
Answer: